function num2Chinese(num) {
	if (!/^\d*(\.\d*)?$/.test(num))
		throw (new Error(-1, "Number is wrong!"));
	var AA = new Array("零", "一", "二", "三", "四", "五", "六", "七", "八", "九");
	var BB = new Array("", "十", "百", "千", "万", "亿", "点");
	var a = ("" + num).replace(/(^0*)/g, "").split("."), k = 0, re = "";
	for ( var i = a[0].length - 1; i >= 0; i--) {
		switch (k) {
		case 4:
			if (!new RegExp("0{4}\\d{" + (a[0].length - i - 1) + "}$")
					.test(a[0]))
				re = BB[4] + re;
			break;
		case 8:
			re = BB[5] + re;
			BB[7] = BB[5];
			k = 0;
			break;
		}
		if (k % 4 == 2 && a[0].charAt(i) == "0" && a[0].charAt(i + 2) != "0")
			re = AA[0] + re;
		if (a[0].charAt(i) != 0)
			re = AA[a[0].charAt(i)] + BB[k % 4] + re;
		k++;
	}
	if (a.length > 1) {
		re += BB[6];
		for ( var i = 0; i < a[1].length; i++) {
			re += AA[a[1].charAt(i)];
		}
	}
	return re;
}